If $x \bigtriangleup y = (8-x)(y)$ and $x \odot y = x^{2}+3y^{2}$, find $(3 \odot -1) \bigtriangleup 4$.
Solution: First, find $3 \odot -1$ $ 3 \odot -1 = 3^{2}+3(-1)^{2}$ $ \hphantom{3 \odot -1} = 12$ Now, find $12 \bigtriangleup 4$ $ 12 \bigtriangleup 4 = (8-12)(4)$ $ \hphantom{12 \bigtriangleup 4} = -16$.